How to Calculate Molarity of Glacial Acetic Acid (99.7% w/w), Alternative Method

Glacial acetic acid is a clear colorless liquid. A 99.7% (w/w) concentrated glacial acetic acid can be obtained from different suppliers. A 99.7% (w/w) glacial acetic acid means that 100 g of glacial acetic acid contains 99.7 g of acetic acid. The density of glacial acetic acid is 1.049 g/ml at 25°C which means that the weight of the 1 ml of glacial acetic acid is 1.049 grams at 25°C. Molarity refers to the number of moles of the solute present in 1 liter of solution. 

Calculation procedure:

Step 1: Since molarity refers to number of moles in 1000 ml solution, calculate the weight of 1000 ml Glacial acetic acid using the following formula:
Formula:

$Density =\frac{weight}{volume}$

or

weight = Density x volume

Weight = 1000 x 1.049 = 1049

Weight of 1000 ml Glacial acetic acid is 1049 g.

Step 2: Calculate how much CH3COOH is in 1049 g of Glacial acetic acid.
100 g Glacial acetic acid contains: 99.7 g of CH3COOH

1 g Glacial acetic acid will contain:  99.7/100 g of CH3COOH

1049 g Glacial acetic acid will contain:  1049 x 99.7/ 100 = 1045.85 g of CH3COOH

Note: Since the volume of 1049 g of Glacial acetic acid is 1000 ml, we can say that 1045.85 g of CH3COOH is present in 1000 ml of Glacial acetic acid.

Step 3: Calculate the number of moles of acetic acid present in 1045.85 grams of acetic acid.
60.05 grams of acetic acid = 1 mole

$1 gram of acetic acid = \frac{1}{60.05 moles}$

$1045.85 grams = \frac{1045.85 x 1}{60.05} = 17.416 moles$

This means 17.416 moles of CH3COOH is present in 1000 ml of Glacial acetic acid.
Therefore, we can say that 1 liter of glacial acetic acid contains 17.416 moles or in other words molarity of glacial acetic acid (99.7% w/w, density 1.049 g/ml) is equal to 17.416.

Use Calculator to calculate the Molarity of concentrated acetic acid (CH3COOH) when concentration is given in % by mass (w/w)

The molecular weight of Acetic acid (CH3COOH): 60.05 g/mol

Author: admin